Forgetting null/base-case handling
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.
Move from brute-force thinking to an efficient approach using tree strategy.
Given the root of a binary tree, return the leftmost value in the last row of the tree.
Example 1:
Input: root = [2,1,3] Output: 1
Example 2:
Input: root = [1,2,3,4,null,5,6,null,null,7] Output: 7
Constraints:
[1, 104].-231 <= Node.val <= 231 - 1Problem summary: Given the root of a binary tree, return the leftmost value in the last row of the tree.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[2,1,3]
[1,2,3,4,null,5,6,null,null,7]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #513: Find Bottom Left Tree Value
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int findBottomLeftValue(TreeNode root) {
Queue<TreeNode> q = new ArrayDeque<>();
q.offer(root);
int ans = 0;
while (!q.isEmpty()) {
ans = q.peek().val;
for (int i = q.size(); i > 0; --i) {
TreeNode node = q.poll();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
}
return ans;
}
}
// Accepted solution for LeetCode #513: Find Bottom Left Tree Value
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func findBottomLeftValue(root *TreeNode) int {
q := []*TreeNode{root}
ans := 0
for len(q) > 0 {
ans = q[0].Val
for i := len(q); i > 0; i-- {
node := q[0]
q = q[1:]
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
}
return ans
}
# Accepted solution for LeetCode #513: Find Bottom Left Tree Value
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
q = deque([root])
ans = 0
while q:
ans = q[0].val
for _ in range(len(q)):
node = q.popleft()
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
return ans
// Accepted solution for LeetCode #513: Find Bottom Left Tree Value
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
pub fn find_bottom_left_value(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut queue = VecDeque::new();
queue.push_back(root);
let mut res = 0;
while !queue.is_empty() {
res = queue.front().unwrap().as_ref().unwrap().borrow_mut().val;
for _ in 0..queue.len() {
let node = queue.pop_front().unwrap();
let mut node = node.as_ref().unwrap().borrow_mut();
if node.left.is_some() {
queue.push_back(node.left.take());
}
if node.right.is_some() {
queue.push_back(node.right.take());
}
}
}
res
}
}
// Accepted solution for LeetCode #513: Find Bottom Left Tree Value
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function findBottomLeftValue(root: TreeNode | null): number {
let ans = 0;
const q = [root];
while (q.length) {
ans = q[0].val;
for (let i = q.length; i; --i) {
const node = q.shift();
if (node.left) {
q.push(node.left);
}
if (node.right) {
q.push(node.right);
}
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.
Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.
Review these before coding to avoid predictable interview regressions.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.